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Chapter 2: Problem 1

Consider the equation \(E=5000+100 n\). a. Find the value of \(E\) for \(n=0,1,20\). b. Express your answers to part (a) as points with coordinates \((n, E)\)

Short Answer

Expert verified
(0, 5000), (1, 5100), (20, 7000)

Step by step solution

01

Identify the Given Equation

The equation provided is E=5000+100n. This equation represents a linear relationship between E and n.
02

Calculate E for n=0

To find E for n=0, substitute n with 0 in the equation: \( E = 5000 + 100 \times 0 \) Thus, \( E = 5000 \) when \( n = 0 \).
03

Calculate E for n=1

To find E for n=1, substitute n with 1 in the equation: \( E = 5000 + 100 \times 1 \) Thus, \( E = 5100 \) when \( n = 1 \).
04

Calculate E for n=20

To find E for n=20, substitute n with 20 in the equation: \( E = 5000 + 100 \times 20 \) Thus, \( E = 7000 \) when \( n = 20 \).
05

Express Answers as Points

Express the calculated values of E as points with coordinates (n, E): For \( n = 0 \), the point is (0, 5000). For \( n = 1 \), the point is (1, 5100). For \( n = 20 \), the point is (20, 7000).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

linear relationships
Linear relationships describe how two variables change together. In a linear equation like E = 5000 + 100n, as n increases, E increases at a constant rate. This constant relationship can be visualized as a straight line when plotted on a graph. Linear equations are straightforward because they have a clear, unchanging rate of change. This is called the slope. For our given equation, the slope is represented by 100. This means for every 1 unit increase in n, E increases by 100.
coordinate points
Coordinate points help us understand the relationship between two variables by showing where they meet on a graph. Each point is defined by an (x, y) pair. In our exercise, n is the x-coordinate and E is the y-coordinate. For the values calculated:
  • When n = 0, we get the point (0, 5000).
  • When n = 1, the point is (1, 5100).
  • And for n = 20, the point is (20, 7000).
These points can be plotted on a graph to visualize the linear relationship between E and n. The resulting graph will show a straight line, affirming our linear equation.
substitution
Substitution is a key method used to find specific values in algebra. Here, to determine the value of E for specific n values, we replace n with 0, 1, or 20 in the equation.E = 5000 + 100n. This process involves:
  • Replacing n in the equation with the given value.
  • Performing the arithmetic to solve for E.
  • For n = 0: Substitute n = 0 to get E = 5000 + 100x0 = 5000.
  • For n = 1: Substitute n = 1 to get E = 5000 + 100x1 = 5100.
  • For n = 20: Substitute n = 20 to get E = 5000 + 100x20 = 7000.
Substitution allows us to pinpoint exact values within an equation, reinforcing our understanding of how variables interact.

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